Problem: $g(t)=t^2-t-42$ 1) What are the zeros of the function? Write the smaller $t$ first, and the larger $t$ second. $\text{smaller }t=$
Explanation: To find the zeros of the function, we need to solve the equation $g(t)=0$. We can do that by factoring $g(t)$. $\begin{aligned} t^2-t-42&=0 \\\\ (t+6)(t-7)&=0 \\\\ t+6=0&\text{ or }t-7=0 \\\\ t={-6}&\text{ or }t={7} \end{aligned}$ There are many ways to find the vertex. We will do it by using the fact that the $t$ -coordinate of the vertex is exactly between the two zeros. $\begin{aligned} \text{vertex's }t\text{-coordinate}&=\dfrac{({-6})+({7})}{2} \\\\ &={\dfrac{1}{2}} \end{aligned}$ Now we can find the vertex's $y$ -coordinate by evaluating $g\left({\dfrac{1}{2}}\right)$ : $\begin{aligned} g\left({\dfrac{1}{2}}\right)&=\left({\dfrac{1}{2}}\right)^2-\left({\dfrac{1}{2}}\right)-42 \\\\ &=\dfrac{1}{4}-\dfrac{1}{2}-42 \\\\ &=-\dfrac{169}{4} \end{aligned}$ In conclusion, $\begin{aligned} \text{smaller }t&=-6 \\\\ \text{larger }t&=7 \end{aligned}$ The vertex of the parabola is at $\left(\dfrac{1}{2},-\dfrac{169}{4}\right)$